Moment of inertia of a circular wire of mass | vedclass

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(B) For a circular ring (or wire) of mass $M$ and radius $R$,the moment of inertia about an axis passing through its center and perpendicular to its p

Vedclass · Accepted Answer

$M R^{2} / 2$

Vedclass · Answer

(B) For a circular ring (or wire) of mass $M$ and radius $R$,the moment of inertia about an axis passing through its center and perpendicular to its plane is $I_{z} = M R^{2}$.According to the perpendicular axis theorem,$I_{z} = I_{x} + I_{y}$,where $I_{x}$ and $I_{y}$ are the moments of inertia about two mutually perpendicular diameters.Since the ring is symmetric,$I_{x} = I_{y} = I_{diameter}$.Therefore,$I_{z} = 2 I_{diameter}$.Substituting the value of $I_{z}$,we get $M R^{2} = 2 I_{diameter}$.Thus,$I_{diameter} = M R^{2} / 2$.

Moment of inertia of a circular wire of mass $M$ and radius $R$ about its diameter is

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